Here we prove that a general disjoint union of lines and +lines has the expected Hilbert function

Note Mat. 35 (2015) no. 1, 1–13. doi:10.1285/i15900932v35n1p1

Postulation of general unions of lines and decorated lines

Edoardo Ballicoⁱ

Department of Mathematics, University of Trento ballico@science.unitn.it

Received: 2.11.2013; accepted: 3.6.2014.

Abstract. A +line A ⊂ P^r, r ≥ 3, is the scheme A = L ∪ v with L a line and v a tangent vector of P^r supported by a point of L, but not tangent to L. Here we prove that a general disjoint union of lines and +lines has the expected Hilbert function.

Keywords:postulation, Hilbert function, lines, zero-dimensional schemes

MSC 2000 classification:primary 14N05, secondary 14H99

Introduction

Fix a line L ⊂ P^r, r ≥ 2, and P ∈ L. A tangent vector of P^r with P as its support is a zero-dimensional scheme Z ⊂ P^r such that deg(Z) = 2 and Zred= {P }. The tangent vector Z is uniquely determined by P and the line hZi spanned by Z. Conversely, for each line D ⊂ P^r with P ∈ D there is a unique tangent vector v with v_red= P and hvi = D. A +line M ⊂ P^r supported by L and with a nilradical at P is the union v ∪ L of L and a tangent vector v with P as its support and spanning a line hvi 6= L. The set of all +lines of P^rsupported by L and with a nilradical at P is an irreducible variety of dimension r − 1 (the complement of L in the (r − 1)-dimensional projective space of all lines of P^r containing P ). Hence the set of all +lines of P^r supported by L is parametrized by an irreducible variety of dimension r. Therefore the set of all +lines of P^r is parametrized by an irreducible variety of dimension 2(r − 1) + r = 3r − 1.

Now assume r ≥ 3. For all integers t ≥ 0 and c ≥ 0 let L(r, t, c) be the set of all disjoint unions X ⊂ P^r of t lines and c +lines. If (t, c) 6= (0, 0), then L(r, t, c) is an irreducible variety of dimension (t + c)(2r − 1) + cr. Fix any X ∈ L(r, t, c) and any integer k > 0. It is easy to check that h⁰(OX(k)) = (k + 1)(t + c) + c and hⁱ(O_X(k)) = 0 for all i > 0 (Lemma 2). A closed subscheme E ⊂ P^r is said to have maximal rank if for every integer k > 0 either h⁰(I_E(k)) = 0 or h¹(IE(k)) = 0, i.e. h⁰(IE(k)) = max{0, ^r+k_r
− h⁰(OE(k))}.

iThis work is partially supported by MIUR and GNSAGA (INDAM) http://siba-ese.unisalento.it/ c 2015 Universit`a del Salento

Theorem 1. Fix integers r ≥ 3, t ≥ 0 and c ≥ 0 such that (t, c) 6= (0, 0). If r ≥ 4, then assume that the characteristic is zero. Then a general X ∈ L(r, t, c) has maximal rank.

We prove Theorem 1 for r = 3 in arbitrary characteristic, while we assume characteristic zero if r ≥ 4. We also get intermediate results (e.g. B_r,k) that may be useful as a sample of lemmas which may be proved with +lines. We see +lines as a tool to prove something involving the Hilbert function of unions of curves and fat points. For an alternative approach to such disjoint unions, see Remark 1.

1 Preliminaries

Remark 1. Fix a line L ⊂ Pⁿ, n ≥ 2, and a linear system V ⊆ H⁰(O_Pⁿ(k)).

Let L⁽¹⁾ be the first infinitesimal neighborhood of L in Pⁿ, i.e. the closed sub- scheme of Pⁿ with (IL)² as its ideal sheaf. Let A be any +line with L as its support. For any closed subscheme B ⊂ Pⁿ set V (−B) := {f ∈ V : f_|B ≡ 0}.

The +line A gives independent conditions to V with the only restriction of that L is the support of L if either V (L) = {0} or dim(V (−A)) = dim(V (−L)) − 1.

A general +lines with L as its supports does not give independent conditions to V with the only restriction that L is its support if and only if V (−L) 6= {0} and V (−L⁽¹⁾) = V (−L). Now assume dim(V (−L⁽¹⁾)) = dim(V (−L)) − γ for some γ > 0. The integer γ is the maximal number of tangent vectors v1, . . . , vγ of Pⁿ supported by points of L and imposing independent conditions to V (−L) (with the restriction that their support is a point of L). So if we only need an integer t + c ≥ 2, t + c disjoint lines and x ≥ 2 tangent vectors supported by some of these lines we may decide to put more than one tangent vector on a single line.

Lemma 1. Let X ⊂ P^r be a closed subscheme such that the nilradical sheaf η ⊆ O_X is supported by finitely many points. Set Y := X_red and fix k ∈ N.

Then:

(1) χ(O_X(k)) = χ(O_Y(k)) + deg(η);

(2) h⁰(I_X(k)) ≤ h⁰(I_Y(k)) ≤ h⁰(I_X(k)) + deg(η);

(3) h¹(I_Y(k)) ≤ h¹(I_X(k)) ≤ h¹(I_Y(k)) + deg(η);

(4) h⁰(I_X(k)) − h¹(I_X(k)) = h⁰(I_Y(k)) − h¹(I_Y(k)) − deg(η).

Proof. By the definition of the reduction of a scheme the sheaf η is the ideal sheaf of Y in X. We have exact sequence (respectively of O_X-sheaves and of O_P^r-sheaves):

0 → η → O_X(k) → O_Y(k) → 0 (1)

0 → I_X(k) → I_Y(k) → η → 0 (2) Since η is supported by finitely many points, we have hⁱ(η) = 0 for all i > 0 and deg(η) = h⁰(η). Use the cohomology exact sequences of (1) and (2). ^QED

Remark 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix A ∈ L(r, t, c), D ∈ L(r, t + c, 0) and set B := A_red.

(1) We have B ∈ L(r, t + c, 0). If A is general in L(r, t, c), then B is general in L(r, t + c, 0).

(2) Assume that D is general in L(r, t + c, 0) and fix a decomposition D = D₁ t D₂ with D₁ ∈ L(r, t, 0) and D₂ ∈ L(r, c, 0). Let E be a general element of L(r, 0, c) with Ered= D2. Then D1 is general in L(r, t, 0), D2

is general in L(r, c, 0) and D₁∪ E is general in L(r, t, c).

Lemma 1 and Remark 2 give the following result.

Lemma 2. Fix integers r ≥ 3, t ≥ 0 and c > 0. Fix X ∈ L(r, t, c) and set Y := Xred. We have Y ∈ L(r, t + c, 0). If A is general in L(r, t, c), then B is general in L(r, t + c, 0). For each integer k > 0 we have h¹(O_X(k)) = 0, h⁰(O_X(k)) = (t + c)(k + 1) + c, h⁰(I_Y(k)) − c ≤ h⁰(I_X(k)) ≤ h⁰(I_Y(k)) and h¹(IY(k)) ≤ h¹(IX(k)) ≤ h¹(IY(k)) + c.

For all integers r ≥ 3 and k ≥ 0 let H_r,k denote the following statement:

Assertion H_r,k, r ≥ 3, k ≥ 0: Fix (t, c) ∈ N² \ {(0, 0)} and take a general X ∈ L(r, t, c). If (k + 1)t + (k + 2)c ≥ ^r+k_k
, then h⁰(IX(k)) = 0. If (k + 1)t + (k + 2)c ≤ ^r+k_k
, then h¹(IX(k)) = 0.

Lemma 3. Fix a general X ∈ L(r, t, c). If 2t + 3c ≤ r + 1, then h¹(I_X(1)) = 0. If 2t + 3c ≥ r + 1, then h⁰(I_X(1)) = 0.

Proof. Since the case c = 0 is obvious, we may assume c > 0 and use induction on c. Fix a general Y ∈ L(r, t, c− 1) and write X = Y t A with A a general +line of P^r. If 2t+3(c−1) ≥ r −1, we immediately see that h⁰(IY ∪Ared(1)) = 0. Hence h⁰(I_X(1)) = 0. Hence we may assume 2t + 3(c − 1) ≤ r − 2. Let M ⊂ P^r be the (2t + 3c − 4)-dimensional linear subspace spanned by Y . Since A is general, it spans a plane N such that M ∩ N = ∅. Hence h⁰(IY ∪A(1)) = h⁰(IY(1)) − 3 =

r + 1 − 2t − 3c. ^QED

Remark 3. Fix an integer r ≥ 3. By the definition of maximal rank and the irreducibility of each L(r, t, c) Theorem 1 is true for the integer r if and only if all Hr,k are true. Since Hr,0 is obviously true, to prove Theorem 1 in P^r it is sufficient to prove H_r,k for all k > 0. Lemma 3 says that H_r,1 is true.

Remark 4. Fix integers r ≥ 3 and k > 0 and suppose you want to prove H_r,k. Fix (t, c) ∈ N²\ {(0, 0)}. First assume t > 0 and (k + 1)(t + c) + c < ^r+k_r
.

Hence (k + 1)(t + c) + (c + 1) ≤ ^r+k_r
. Suppose that h¹(I_X(k)) = 0 for a general X ∈ L(r, t−1, c+1). Then h¹(IY(k)) = 0 for a general Y ∈ L(r, t, c) (Lemma 1).

Now assume c > 0 and (k + 1)(t + c) + c > ^r+k_r
and so (k + 1)(t + c) + (c − 1) ≥

r+k

r
. Suppose that h⁰(IA(k)) = 0 for a general A ∈ L(r, t + 1, c − 1). Then h⁰(I_B(k)) = 0 for a general B ∈ L(r, t, c). Therefore to prove H_r,k it is sufficient to test all (t, c) such that either (k + 1)(t + c) + c = ^k+r_r

or t = 0 and (k + 2)c < ^k+r_r

or c = 0 and (k + 1)t > ^r+k_r
. We do not need to test the pairs (t, 0) by [6]. Among the pairs (0, c) with (k + 2)c ≤ ^r+k_r
it is sufficient to test the ones with ^r+k_r
− k − 1 ≤ (k + 2)c ≤ ^r+k_r
.

For all integers r ≥ 3 and k ≥ 0 define the integers mr,k and nr,k by the relations

(k + 1)m_r,k+ n_r,k=r + k k



, 0 ≤ n_r,k≤ k (3)

Remark 5. Fix integers r ≥ 3 and k > 0. Since m_r,k ≤ k and k(k + 1) ≤

k+3

3
≤ ^r+k_r
, we get m_r,k≥ n_r,k.

For all integers r ≥ 3 and k ≥ 0 set ur,k := d ^r+k_r
/(k + 2)e and v_r,k :=

(k + 2)u_r,k− ^r+k_r
.

Notice that

(k + 2)(u_r,k− v_r,k) + (k + 1)v_r,k=r + k r



(4) and that 0 ≤ v_r,k≤ k + 1.

For all integers k > 0 let Ar,k denote the following assertion:

Assertion A_r,k, k > 0: Let X ⊂ P^r be a general union of v_r,k lines and ur,k− v_r,k +lines. Then h⁰(IX(k)) = 0.

2 The proof in P³

In this section we prove the case r = 3 of Theorem 1.

Lemma 4. Fix integers a ≥ 0, b ≥ 0, y ≥ 0. Let Z ⊂ Q be a general union of y tangent vectors. Then h⁰(I_Z(a, b)) = max{0, (a + 1)(b + 1) − 2y} and h¹(I_Z(a, b)) = max{0, 2y − (a + 1)(b + 1)}.

Proof. By the semicontinuity theorem for cohomology ([5], III.12.8) it is suffi- cient to find a disjoint union W ⊂ Q of y tangent vectors such that h⁰(IW(a, b)) = max{0, (a + 1)(b + 1) − 2y}. It is obviously sufficient to do it for the integers

y = b(a+1)(b+1)c and y = d(a+1)(b+1)/2e. First assume a odd. Let L₀, . . . , L_b be b+1 distinct lines of type (0, 1). Let E_i⊂ L_ibe any disjoint union of (a+1)/2 tangent vectors. In this case we may take W = E1∪ · · · ∪ E_b. In the same way we conclude if b is odd. Hence we may assume that both a and b are even. If b = 0, then take y tangent vectors of L₀. Similarly we conclude if a = 0. Hence we may assume a ≥ 2 and b ≥ 2 and use induction on a. It is obviously suffi- cient to check the integers y such that 2y ≥ (a + 1)(b + 1) − 1. Fix a smooth C ∈ |O_Q(2, 2)|. C is a smooth elliptic curve and in particular it is irreducible.

Take a general S ⊂ C with ](S) = a+b. Let W ⊂ C be the union of the 2-points of C with the points of S as their support, i.e. the degree 2a+2b effective divisor of C in which each point of S appears with multiplicity two. Let W⁰ ⊂ Q be a union of y − a − b general tangent vectors. Set Z := W ∪ W⁰. By the inductive assumption we have h⁰(I_W⁰(a − 2, b − 2)) = max{0, (a − 1)(b − 1) − 2y + 2a + 2b}, i.e. h⁰(I_W⁰(a − 2, b − 2)) = max{0, (a + 1)(b + 1) − 2y} and h¹(I_W⁰(a − 2, b − 2)) = max{0, 2y − (a + 1)(b + 1)}. There are only finitely many (four in characteris- tic 6= 2, one or two in characteristic 2) line bundles R with R^⊗2 ∼= OC(a, b).

Since C has genus > 0 for general S the line bundle O_C(S) is not one of them.

Hence W /∈ |O_C(a, b)|. Since deg(W ) = deg(OC(a, b)), Riemann-Roch gives hⁱ(C, O_C(a, b)(−W )) = 0, i = 0, 1. Since Res_C(Z) = W⁰, the Castelnuovo’s sequence gives hⁱ(I_Z(a, b)) = hⁱ(I_W⁰(a − 2, b − 2)). ^QED

We have u3,k := d(k + 3)(k + 1)/6e and v3,k := (k + 2)u3,k − ^3+k₃
. Write k = 6m + b with 0 ≤ b ≤ 5. We have u_3,6m = 6m²+ 4m + 1, v_3,6m = 3m + 1, u_3,6m+1= 6m²+ 6m + 2, v_3,6m+1= 4m + 2, u_3,6m+2= 6m²+ 8m + 3, v_3,6m+2= 3m + 2, u3,6m+3 = 6m²+ 10m + 4, v3,6m+3 = 0, u3,6m+4 = 6m² + 12m + 6, v3,6m+4 = m + 1, u3,6m+5 = 6m² + 14m + 8, v3,6m+5 = 0. The construction below works (in particular Lemma 6) only because u_3,6m+7−v_3,6m+7≥ u_3,6m+5− v3,6m+5 (both sides of the inequality are equal to u3,6m+5 = 6m² + 14m + 8).

Without this inequality we would have needed a longer proof. In general we need u_3,k+2− v_3,k+2≥ u_3,k− v_3,k for all k > 0, but only in the case k = 6m + 5 the right hand side is not much bigger than the left hand side.

Let Q ⊂ P³ be a smooth quadric surface. We have Pic(Q) ∼= Z² and we take two distinct, but intersecting, lines contained in Q as a basis of Pic(Q). We will call |OQ(1, 0)| and |OQ(0, 1)| the two rulings of Q and call any D ∈ |OQ(a, b)| a divisor (or a curve) of type (a, b). Fix a line L ⊂ Q, P ∈ L and let γ be the set of all +lines A ⊂ P³ with L as their support and P as the support of the nilradical sheaf of OA. The set γ is the complement of a point in a two-dimensional projective space (it is P(TPP³) \ P(TPL)). The set γ⁰ of all A ∈ γ contained in Q is the line P(TPQ) minus the point P(TPL). If A ∈ γ⁰, then A ⊂ Q and ResQ(A) = ∅. If A /∈ γ⁰, then A ∩ Q = L (as schemes) and ResQ(A) = {P } (as schemes). Now take A ⊂ Q, see L as a divisor of Q; we have Res_L(A) = {P }.

Lemma 5. A_r,1 and A_3,2 are true.

Proof. Ar,1 is is true by Remark 3.

We have v_3,2= 2 and u_3,2− v_3,2= 1. Take any B = L₁t L₂t L₃ ∈ L(3, 3, 0).

B is contained in a unique quadric surface, Q⁰, and Q⁰ is a smooth quadric. Let A ⊂ P³be a general +line with L3as its support. We have L1∪L₂∪A ∈ L(3, 2, 1).

Since A * Q⁰, we have h⁰(I_L1∪L₂∪A(2)) = 0. ^QED Of course, A_r,k makes sense only if u_r,k− v_r,k ≥ 0; this is the reason why we didn’t defined A_r,0. By the semicontinuity theorem for cohomology ([5], III.12.8) to prove A3,k it is sufficient to find A ∈ L(3, v3,k, u3,k − v_3,k) such that h⁰(I_A(k)) = 0. Fix an integer k > 0 and let X ⊂ P³ be a general union of v_3,k lines and u_3,k− v_3,k +lines. We have h⁰(O_X(k)) = (u_3,k− v_3,k)(k + 2) + v_3,k(k + 1) = ^k+3₃
, the latter equality being true by the definition of the integer v_3,k. Hence h¹(IX(k)) = h⁰(IX(k)).

Lemma 6. A_3,k ⇒ A_3,k+2 for all k > 0.

Proof. We have 0 ≤ u_3,k+2− u_3,k ≤ k + 2.

(a) First assume v3,k+2 ≥ v_3,k, i.e. k ≡ 0, 3, 4, 5 (mod 6). Notice that in all cases we have u_3,k− v_3,k ≤ u_3,k+2− v_3,k+2(we even have equality if k = 6m + 5, because u_3,6m+7= 6m²+ 18m + 14, v_3,6m+7= 4m + 6, u_3,6m+5= 6m²+ 14m + 8 and v3,6m+5 = 0). Let L1 ⊂ Q be the union of v_3,k+2− v_3,k distinct lines of type (1, 0) and L2 ⊂ Q the union of (u_3,k+2− v_3,k+2) − (u_3,k − v_3,k) distinct lines of type (1, 0) with L₁∩ L₂ = ∅. Let A₂ ⊂ Q be the union of (u_3,k+2− v_3,k+2)−(u_3,k−v_3,k) general +lines contained in Q and with the lines of L₂as its support. Let S2⊂ L₂be the support of the nilradical of OA2. Take a general Y ∈ L(3, v_3,k, u_3,k−v_3,k). For general Y we have Y ∩(L₁∪L₂) = ∅ and so Y ∪A₂∪L₁ ∈ L(3, v_3,k+2, u_3,k+2− v_3,k+2). By the semicontinuity theorem for cohomology ([5], III.12.8) it is sufficient to prove h⁰(IY ∪A2∪L1(k)) = 0. Since ResQ(Y ∪A2∪L₁) = Y and h⁰(I_Y(k −2)) = 0, it is sufficient to prove h⁰(Q, I_{Q∩(Y ∪A2∪L1)}(k +2)) = 0.

Since L₁∪ L₂ ⊂ (Y ∩ Q) ∪ A₂ ∪ L₁ and L₁∪ L₂ ∈ |O_Q(u_3,k+2− u_3,k, 0)|, it is sufficient to prove h⁰(Q, IRes_L1∪L2((Y ∩Q)∪A2∪L1)(k +2−u3,k+2+u3,k, k +2)) = 0.

We have ResL1∪L2((Y ∩ Q) ∪ A2∪ L₁) = (Y ∩ Q) ∪ S2. For general A2 the set S₂ is a set containing a general point of (u_3,k+2− v_3,k+2) − (u_3,k− v_3,k) general lines of type (1, 0) and nothing else. Hence S₂ may be considered as a general union of (u3,k+2 − v_3,k+2) − (u3,k − v_3,k) points of Q. For general Y the set Y ∩ Q is a general subset of Q with cardinality 2u_3,k. Hence it is sufficient to check that ]((Y ∩ Q) ∪ S₂) = h⁰(Q, O_Q(k + 2 − u_3,k+2+ u_3,k, k + 2)), i.e.

2u3,k + (u3,k+2− v_3,k+2) − (u3,k − v_3,k) = (k + 3 − u3,k+2 + u3,k)(k + 3), i.e.

2u_3,k+ (k + 2)(u_3,k+2− u_3,k) = (k + 3)²+ v_3,k+2− v_3,k. Taking the difference of

(4) for the integer k⁰ = k + 2 from (4) and using that ^k+5₃
− ^k+3₃
= (k + 3)² we get 2u_3,k + (k + 2)(u_3,k+2− u_3,k) = (k + 3)²+ v_3,k+2− v_3,k, as wanted.

(b) Now assume v_3,k+2 < v_3,k, i.e. k ≡ 1, 2 (mod 6). Take a general Y ∈ L(3, v_3,k, u_3,k− v_3,k) and write Y = E ∪ F with E ∈ L(3, v_3,k+2, u_3,k− v_3,k) and F ∈ L(3, v3,k−v_3,k+2, 0). For general Y we have h⁰(IY(k)) = 0 (by the inductive assumption) and Y ∩ Q is a general subset of Q with cardinality 2u_3,k. For each line L ⊆ F fix one of the point P_L∈ L∩Q and call v_La general tangent vector of Q at PL. Let AL= L ∪ vL⊂ P³be the +lines with L as its reduction, PLas the support of its nilradical and containing v_L. Set G := ∪_L∈FA_L. Let M ⊂ Q be a union of u_3,k+2− u_3,k general lines of type (1, 0). Let N ⊂ Q be a general union of u3,k+2−u_3,k+lines with M as the union of their support. We have E ∪G∪N ∈ L(3, v_3,k+2, u_3,k+2− v_3,k+2). Since Res_Q(E ∪ G ∪ N ) = Y and h⁰(I_Y(k)) = 0, it is sufficient to prove h⁰(Q, I_{Q∩(E∪G∪N )}(k + 2, k + 2)). Since M ⊂ Q ∩ (E ∪ G ∪ N ), it is sufficient to prove h⁰(Q, IResM(Q∩(E∪G∪N ))(k +2−u3,k+2+u3,k, k +2)) = 0.

The scheme G ∩ Q is a general union δ of v_3,k tangent vectors of Q and a general union of v_3,k − v_3,k+2 points of Q; we do not want to use here that general tangent vectors gives the maximal possible number of conditions to any linear system, because it requires characteristic zero ([4], [1], Lemma 1.4); however, since v_3,k ≤ 2(k+2)/3, it is obvious that h¹(Q, I_δ(k+2−u_3,k+2+u_3,k, k+2)) = 0;

alternatively, use Lemma 4. Set S := ResM(N ). The set S contains one point for each line of M and it is general with this condition. Since M is a general union of u_3,k+2− u_3,k lines of type (1, 0), S may be considered as a general subset of Q with its cardinality. The set E ∩ Q is a general subset of Q with cardinality 2u_3,k− 2v_3,k+2. Since 2(v_3,k− v_3,k+2) + (v_3,k− v_3,k−2) + (u_3,k+2− u_3,k) + 2(u_3,k− v_3,k+2) = (k + 3 − u_3,k+2+ u_3,k)(k + 3), we are done. ^QED

Lemma 7. For all integers k > 0 and c > 0 such that c(k + 2) ≤ ^k+3₃
we have h¹(I_X(k)) = 0 for a general X ∈ L(3, 0, c).

Proof. If k = 1, then c = 1. The lemma is obvious in this case.

Now assume k = 2. It is sufficient to do the case c = 2. Take A = A1∪ A₂ ∈ L(3, 0, 2) with L₁ and L₂ two different lines of type (1, 0) of Q, A₁ ⊂ Q and general with these restrictions, A₂ * Q and general among the +lines supported by Q. We get h⁰(Q, OQ∩A(2)) = 2 and h⁰(IResQ(A)) = 0, because ResQ(A) 6= ∅.

From now on we assume k ≥ 3. Lemmas 5 and 6 give that A_3,k−2 and A_3,k are true. We have c ≤ u_3,k and c ≤ u_3,k − 1 if v_3,k > 0. If c ≤ u_3,k − v_3,k, then we may use A3,k. In particular we are done if v3,k = 0. Hence we may assume v_3,k > 0. In this case it is sufficient to do the case c = u_3,k − 1. Fix a general Y ∈ L(3, v_3,k−2, u_3,k−2− v_3,k−2). We have hⁱ(I_Y(k − 2)) = 0, i = 0, 1, by A3,k−2. We mimic part (b) of the proof of Lemma 6. Write Y = E t F with E ∈ L(3, 0, u_3,k − v_3,k) and F ∈ L(3, v_3,k, 0). For each line L ⊆ F fix

one of the point P_L ∈ L ∩ Q and call v_L a general tangent vector of Q at P_L. Let A_L = L ∪ v_L ⊂ P³ be the +line with L as its reduction, P_L as the support of its nilradical and containing vL. Set G := ∪L∈FAL. Let M ⊂ Q be a union of u_3,k+2− u_3,k− 1 general lines of type (1, 0). Let N ⊂ Q be a general union of u_3,k+2 − u_3,k − 1 +lines with M as the union of their support. Take X := E ∪ G ∪ N ∈ L(3, 0, u3,k− 1). Since v_3,k ≤ k + 1, as in part (b) of the proof of Lemma 6 we get h¹(Q, I_X∩Q(k)) = 0 and hence h¹(I_X(k)) = 0. ^QED

Proof of Theorem 1 for r = 3: It is sufficient to prove H_3,k for all k ≥ 2 (Remark 3). Fix an integer k > 0. It is sufficient to check the Hilbert function in degree k of a general element of L(3, t, c) with either t = 0 and (k + 2)c ≤ ^k+3₃
or (k + 1)t + (k + 2)c = ^k+3₃

(Remark 4). By Lemma 7 it is sufficient to check the pairs (t, c) with t > 0 and (k + 1)t + (k + 2)c = ^k+3₃
, i.e. (since the integers k + 1 and k + 2 are coprime) the pairs (t, c) with t = v_3,k + (k + 2)α, c = u_3,k− v_3,k− (k + 1)α for some non-negative integer α such that (k + 1)α ≤ u3,k− v_3,k. By [6] we may assume c > 0. By Lemma 6 we may assume t > v3,k. Since v3,2= 2, u3,2= 3 and A3,2 is true (Lemma 5), we may assume k ≥ 3. By induction on k we may assume that hⁱ(I_W(k − 2)) = 0, i = 0, 1, for a general W ∈ L(3, t⁰, c⁰) for all non-negative integers t⁰, c⁰such that (k−1)t⁰+kc⁰ = ^k+1₃
.

Fix a general Y ∈ L(3, m_3,k−2− n_3,k−2, n_3,k−2). We have hⁱ(IY(k − 2)) = 0, i = 0, 1.

Claim 1: We have t + c ≥ m_3,k−2.

Proof of Claim 1: Assume t + c ≤ m_3,k−2− 1, i.e. assume (k − 1)(t + c) + k − 1 ≤ ^k+1₃
. Since (k + 1)t + (k + 2)c = ^k+3₃
, we get (k + 2)/(k − 1) >

k+3

3
/ ^k+1₃
= (k + 3)(k + 2)/k(k − 1), a contradiction.

Claim 2: We have c ≥ n_3,k−2.

Proof of Claim 2: If k − 2 ≡ 0, 1 (mod 3), then n_3,k−2 = 0. If k − 2 ≡ 2 (mod 3), then n_3,k−2= (k − 1)/3. Hence we may assume k ≡ 4 (mod 3). Since n_3,k = 0, c > 0 and (k + 1)t + (k + 2)c = ^k+3₃
, we get c = β(k + 1) for some integer β > 0. Hence c ≥ k + 1 > n_3,k−2.

Notice that m_3,k−2 ≥ n_3,k−2. Fix a general Y ∈ L(3, m_3,k−2−n_3,k−2, n_3,k−2).

By the inductive assumption we have hⁱ(IY(k − 2)) = 0, i = 0, 1. Set e :=

t + c − m_3,k−2. Claim 1 gives e ≥ 0. Take a general union M ⊂ Q of e lines of type (1, 0).

(a) Assume c − n_3,k−2 ≤ e. Claim 2 gives c − n_3,k−2 ≥ 0. Write M = M1t M₂ with M2 a union of c − n3,k−2 lines and M1 a union of e − (c − n3,k−2) lines. Let A₂ ⊂ Q be a general union of c − n_3,k−2 +lines with the lines in M₂ as their support. Let S₂ be the support of the nilpotent sheaf of A₂. Since A2 is general, S2 is obtained taking for each line L ⊆ M2 a general point of L. Set X := Y ∪ M₁∪ A₂. Since X ∈ L(3, t, c), it is sufficient to prove that

h¹(I_X(k)) = 0. Since Res_Q(X) = Y and h¹(I_Y(k − 2)) = 0, it is sufficient to prove that h¹(Q, I_X∩Q(k)) = 0. Since Res_M(X ∩Q) = (Y ∩Q)∪S₂, it is sufficient to prove that h¹(Q, I_{(Y ∩Q)∪S2}(k − e, k)) = 0. Since M2 is general and for each line L ⊂ M₂the set S₂∩L is a general point of L, S₂is a general subset of Q with cardinality c − n_3,k−2. Since Y is general, the set (Y ∩ Q) ∪ S₂ is a general subset of Q of cardinality 2m_3,k−2+ c − n_3,k−2. Since (k − 1)m_3,k−2+ n_3,k−2 = ^k+1₃
, (k + 1)t + (k + 2)c = ^k+3₃
, ^k+3₃
− ^k+1₃
= (k + 1)² and e = t + c − m_3,k−2, we have 2m3,k−2+ (k + 1)e + c − n3,k−2 = (k + 1)², i.e. ](S2 ∪ (Y ∩ Q)) = (k + 1)(k + 1 − e) = h⁰(Q, O_Q(k − e, k)). Hence hⁱ(Q, I_{(Y ∩Q)∪S2}(k − e, k)) = 0, i = 0, 1.

(b) Now assume c − n_3,k−2 > e. For each line R ⊆ M fix a general OR ∈ R and call v_R a general tangent vector of Q with OR as its support.

Set R⁺ := R ∪ vR, M⁺ := ∪R⊆MR⁺ and S := ∪R⊆MOR. Since M is general and each O_R is general in R, S is a general subset of Q with cardinality e.

We have M⁺ ⊂ Q and hence M⁺∩ Q = M⁺ and ResQ(M⁺) = ∅. Set g :=

c − n_3,k−2− e. Since e = t + c − m_3,k−2, we get m_3,k−2− n_3,k−2= g + t > t. Write Y = Y₁ t Y₂ with Y₂ ∈ L(3, 0, n_3,k−2) and Y₁ ∈ L(3, m_3,k−2− n_3,k−2, 0). Since m3,k−2− n_3,k−2= g + t ≥ t, we may write Y1 = Y3t Y₄ with Y3 ∈ L(3, t, 0) and Y4 ∈ L(3, g, 0). For each line L ⊆ Y₄ fix one of the two points, say OL, of L ∩ Q and let w_L be a general tangent vector of Q with O_L as its support; set L⁺:=

L ∪ w_L∈ L(3, 0, 1). Set Y₄⁺:= ∪_L⊆Y4L⁺ and X⁰ := M⁺∪ Y₄⁺∪ Y₂∪ Y₃. Since X⁰ ∈ L(3, t, c), it is sufficient to prove that h¹(I_X⁰(k)) = 0. Since ResQ(X⁰) = Y and h¹(I_Y(k−2)) = 0, by the Castelnuovo’s sequence it is sufficient to prove that h¹(Q, I_X⁰_∩Q(k)) = 0. The scheme X⁰∩ Q is the union of M⁺and (Y₃∪ Y₄⁺) ∩ Q.

We have ResM(M⁺∪ (Y₃∪ Y₄⁺) ∩ Q) = S ∪ ((Y \ Y4) ∩ Q) ∪S

L⊆Y4wL. Since Y ∩ Q is a general subset of Q with cardinality 2m_3,k−2, the scheme Z :=

S ∪ ((Y \ Y₄) ∩ Q) ∪S

L⊆Y4w_L is a general union of g tangent vectors of Q and e + 2m3,k−2− g points of Q. By the Castelnuovo’s sequence it is sufficient to prove that h¹(Q, I_Z(k − e, k)) = 0. By Lemma 4 it is sufficient to check that deg(Z) ≤ (k − e + 1)(k + 1). By (3) for the integers r = 3 and k⁰= k − 2 and the equality (k + 1)t + (k + 2)c = ^k+3₃
we get deg(Z) = (k + 1)(k − e + 1). ^QED

3 When r > 3

In this section we prove Theorem 1 for all integers r ≥ 4. For numerical reasons this is easier than in the case r = 3 (as it was in [6] and [2]). The proof in characteristic zero is very short and we will only give it.

Lemma 8. For all integers r ≥ 3 and k ≥ 2 we have m_r,k−1< u_r,k.

Proof. We have km_r,k−1≤ ^k+r−1_r
and (k + 2)u_r,k≥ ^r+k_r
. Note that

r + k r

.r + k − 1 r



= (r + k)/k

and that (r + k)k > (k + 2)k for all r ≥ 3. ^QED We need the following assumption B_r,k:

Br,k, r ≥ 4, k > 0. Fix a hyperplane H ⊂ P^r. There is X ∈ L(r, mr,k− n_r,k, n_r,k) such that the support of the nilradical sheaf of X is contained in H and h⁰(I_X(k)) = 0.

For all X ∈ L(r, m_r,k − n_r,k, n_r,k) we have h⁰(O_X(k)) = ^r+k_r

and so h¹(I_X(k)) = h⁰(I_X(k)).

Lemma 9. For all integers r ≥ 4 and k ≥ 2 we have mr,k≥ m_r,k−1. Proof. We have

m_r,k−1+ (k + 1)(m_r,k− m_r,k−1) + n_r,k− n_r,k−1 =r + k − 1 r − 1



(5) Assume m_r,k≤ m_r,k−1− 1. Since n_r,k− n_r,k−1≤ k, (6) gives

m_r,k−1− 1 ≥r + k − 1 r − 1



Since km_r,k−1 ≤ ^r+k−1_r
and k ^r+k−1_r−1
= r ^r+k−1_r
, we get −k ≥ (r − 1) ^r+k−1_r−1
,

a contradiction. ^QED

Lemma 10. Fix an integer r ≥ 4 and assume that Theorem 1 is true in P^r−1. Then B_r,k is true for all k > 0.

Proof. Br,1is true by Remark 3. Hence we may assume k ≥ 2 and that B_r,k−1is true. Fix Y ∈ L(r, m_r,k−1−n_r,k−1, n_r,k−1) such that the support of the nilradical sheaf of Y is contained in H and h⁰(I_Y(k − 1)) = 0. By the semicontinuity theorem for cohomology ([5], III.12.8) we may assume that Y is general among the elements of L(r, m_r,k−1− n_r,k−1, n_r,k−1) whose nilradical sheaf is supported by points of H. Hence we may assume that no irreducible component of Y_red is contained in H, that Yred is a general subset of H with cardinality mr,k−1

and that for each +line A ⊂ Y , say A = L ∪ v_L, the tangent vector v_L of A is not contained in H. The latter assumption implies Y_red∩ H = Y ∩ H (scheme- theoretic intersection) and ResH(Y ) = Y . We have mr,k ≥ m_r,k−1 (Lemma 9).

(a) In this step we assume n_r,k < n_r,k−1. Let F ⊂ H be a general union of m_r,k − m_r,k−1 lines, with the only restriction that exactly n_r,k−1− n_r,k of them contain a point of Y_red∩ H. We have m_r,k−1− n_r,k−1− (n_r,k−1− n_r,k) + m_r,k− m_r,k−1 = m_r,k− 2n_r,k−1 + n_r,k. The scheme Y ∪ F is a disjoint union of n_r,k−1− n_r,k sundials, n_r,k +lines and m_r,k − 2n_r,k−1 + n_r,k lines. Since a sundial is a flat limit of a family of elements of L(r, 2, 0) ([2]), it is sufficient to prove h⁰(I_{Y ∪F}(k)) = 0. Since the set Y_red∩ H is general in H, F may be considered as a general union of lines. Hence F has maximal rank. By (5) we have h¹(H, IF(k)) = 0 and h⁰(H, IF(k)) = mr,k−1 − n_r,k−1 + nr,k. Since for fixed Y_red∩ H ∩ F we may deform the other components of Y so that the other m_r,k−1 − n_r,k−1 + n_r,k points of Y_red∩ (H \ F ) are general in H, then hⁱ(H, I_{H∩(Y ∪F )}(k)) = 0. Castelnuovo’s sequence gives h⁰(IY ∪F(k)) = 0.

(b) In this step we assume n_r,k ≥ n_r,k−1 and m_r,k− n_r,k ≥ m_r,k−1 − n_r,k−1. Let E ⊂ H be a general union of m_r,k− n_r,k− (m_r,k−1− n_r,k−1) lines and n_r,k − n_r,k−1 +lines. We have Y ∪ E ∈ L(r, m_r,k − n_r,k, n_r,k) and the support of the nilradical sheaf of Y ∪ E is contained in H. By (5) we have h⁰(E, OE(k)) + deg(Y ∩ H) = ^r+k−1_r−1
. Since Theorem 1 is true in P^r−1, we have h¹(H.IE(k)) = 0. Since Y ∩ H is a general union of m_r,k−1 points of H, (5) implies hⁱ(H, I_{(Y ∪E)∩H}(k)) = 0.

(c) In this step we assume n_r,k≥ n_r,k−1and m_r,k− n_r,k< m_r,k−1− n_r,k−1. Therefore g := n_r,k−n_r,k−1−(m_r,k−m_r,k−1) > 0. Since n_r,k≤ k, we have g ≤ k.

Since km_r,k−1+n_r,k−1 = ^r+k−1_r
and n_r,k−1≤ k−1, we have g ≤ m_r,k−1−n_r,k−1. Take a general union G ⊂ H of mr,k−m_r,k−1+lines. Since Theorem 1 is assumed to be true in P^r−1, G has maximal rank. By (5) we have h¹(H, I_G(k)) = 0 and h⁰(H, I_G(k)) = m_r,k−1+ g. Write Y = Y₁t Y₂t Y₃ with Y₃ ∈ L(r, 0, n_r,k−1), Y1 ∈ L(r, m_r,k−1−n_r,k−1, 0) and Y2 ∈ L(r, g, 0). For each line L ⊆ Y₂let vLbe the general tangent vector of H with L ∩ H as its support. Set A2 := ∪L⊆Y2(L ∪ vL).

Since U := Y₁∪ A₂∪ Y₃∪ G ∈ L(r, m_r,k− n_r,k, n_r,k), it is sufficient to prove that h⁰(IU(k)) = 0. We have ResH(U ) = Y , because each vLis contained in H. The scheme U ∩ H is the union of G, m_r,k−1− n_r,k−1− g general points of H and g general tangent vectors of H. Hence hⁱ(H, I_{U ∩H}(k)) = 0, i = 0, 1 ([1], Lemma

1.4). ^QED

Proof of Theorem 1 for r > 3: Let H ⊂ P^r be a hyperplane. We use induction on r, the starting case being the one with r = 3 proved in section 2.

Hence we assume Theorem 1 in H ∼= P^r−1 for all L(r − 1, t⁰, c⁰). By Remark 3 it is sufficient to prove H_r,k for all k > 0. Hr,1 is true (Remark 3). Hence we may assume k ≥ 2 and that H_r,k−1 is true. By Remark 4 it is sufficient to prove H_r,k for the pairs (t, c) such that either t = 0 and ^r+k_r
− k − 1 ≤ c(k + 2) ≤ ^r+k_r
or t(k + 1) + (k + 2)c = ^r+k_r

and c > 0. If ^r+k_r
− k − 1 ≤ c(k + 2) ≤